chapter9 - Chapter 9 Solutions to §9.1 1 F(s = ∠∞ e...

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Unformatted text preview: Chapter 9 Solutions to Β§9.1 1. F(s) = ∫ ∞ e –st e 2t dt = lim n β†’βˆž ∫ n e (2–s)t dt = 1 2 – s lim n β†’βˆž [ ] e (2–s)t n = 1 s – 2 if s > 2. 3. F(s) = ∫ ∞ e –st sin btdt = lim n β†’βˆž ∫ n e –st sin btdt = lim n β†’βˆž e –st [–sin bt – bcos bt] s 2 + b 2 n = b s 2 + b 2 if s > 0. 5. F(s) = ∫ ∞ e –st cosh btdt = lim n β†’βˆž ⌑ ⌠ n e –st e bt + e –bt 2 dt = lim n β†’βˆž e (b – s)t 2(b – s) – e –(b + s)t 2(b + s) n = lim n β†’βˆž e (b – s)n – 1 2(b – s) – e –(b + s)n + 1 2(b + s) = s s 2 – b 2 where s > |b|. 7. F(s) = ∫ ∞ e –st (2t)dt = 2lim n β†’βˆž ∫ n e –st tdt = 2lim n β†’βˆž e –st (–st – 1) s 2 n = 2 lim n β†’βˆž –n se sn – 1 s 2 e sn + 1 s 2 = 2 s 2 where s > 0. 9. F(s) = ∫ 2 e –st (1)dt + ∫ 2 ∞ e –st (–1)dt = e –st –s 2 + lim n β†’βˆž ∫ 2 n (–e –st )dt = [ e –2s –s + 1 s ] + lim n β†’βˆž e –st s 2 n = 1 s – 1 e 2s s + lim n β†’βˆž [ 1 se sn – 1 se 2s ] = 1 s (1 – 2e –2s ) provided s > 0. 11. F(s) = ∫ ∞ e –st (e t sin t)dt = lim n β†’βˆž ∫ n e (1 – s)t sin tdt = lim n β†’βˆž e (1 – s)t [sin t – cos t] (s – 1) 2 + 1 n = lim n β†’βˆž e (1 – 2)n [sin n – cos n] (s – 1) 2 + 1 + 1 (s – 1) 2 + 1 = 1 (s – 1) 2 + 1 where s > 1. 267 268 Ch. 9 The Laplace Transform and Some Elementary Applications 13. L[2t – e 3t ] = L[2t] – L[e 3t ] = 2L[t] – 1 s – 3 = 2( 1 s 2 ) – 1 s – 3 = 2 s 2 – 1 s – 3 , where s > 3. 15. L[cosh bt] = L e bt 2 + e –bt 2 = 1 2 L[e bt ] + 1 2 L[e –bt ] = 1 2 1 s – b + 1 2 1 s + b = s s 2 – b 2 where s > |b|. 17. L[3t 2 – 5cos 2t + sin 3t] = 3L[t 2 ] – 5L[cos 2t] + L[sin 3t] = 3 2! s 3 – 5 s s 2 + 4 + 3 s 2 + 9 = 6 s 3 – 5s s 2 + 4 + 3 s 2 + 9 . 19. L[2e –3t + 4e t – 5sin t] = 2L[e –3t ] + 4L[e t ] – 5L[sin t] = 2 s + 3 + 4 s – 1 – 5 s 2 + 1 where s > 1. 21. L[4cos 2 bt] = 2L[2cos 2 bt] = 2L[cos 2bt + 1] = 2L[cos 2bt] + 2L[1] = 2 ( s 4b 2 + s 2 ) + 2 ( 1 s ) = 4(s 2 + 2b 2 ) (s 2 + 4b 2 )s where s > 0. 23. The graph of f is piece–wise continuous on [0, ∞ ). 25. The graph of f is not piece–wise continuous on [0, ∞ ), because lim n β†’1 + f(t) = ∞ , i.e. the right– hand limit as t approaches 1 from above is not finite. 27. The graph of f is piece–wise continuous on [0, ∞ )....
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chapter9 - Chapter 9 Solutions to §9.1 1 F(s = ∠∞ e...

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